The digits of some specific integers permute or shift cyclically when they are multiplied by a number n. Examples are:
These specific integers, known as transposable integers, can be but are not always cyclic numbers. The characterization of such numbers can be done using repeating decimals (and thus the related fractions), or directly.
For any integer coprime to 10, its reciprocal is a repeating decimal without any non-recurring digits. E.g. 1⁄143 = 0.006993006993006993...
While the expression of a single series with vinculum on top is adequate, the intention of the above expression is to show that the six cyclic permutations of 006993 can be obtained from this repeating decimal if we select six consecutive digits from the repeating decimal starting from different digits.
This illustrates that cyclic permutations are somehow related to repeating decimals and the corresponding fractions.
The greatest common divisor (gcd) between any cyclic permutation of an m-digit integer and 10^{m} − 1 is constant. Expressed as a formula,
where N is an m-digit integer; and N_{c} is any cyclic permutation of N.
For example,
gcd(091575, 999999) = gcd(3^{2}×5^{2}×11×37, 3^{3}×7×11×13×37) = 3663 = gcd(915750, 999999) = gcd(157509, 999999) = gcd(575091, 999999) = gcd(750915, 999999) = gcd(509157, 999999)
If N is an m-digit integer, the number N_{c}, obtained by shifting N to the left cyclically, can be obtained from:
where d is the first digit of N and m is the number of digits.
This explains the above common gcd and the phenomenon is true in any base if 10 is replaced by b, the base.
The cyclic permutations are thus related to repeating decimals, the corresponding fractions, and divisors of 10^{m}−1. For examples the related fractions to the above cyclic permutations are thus:
Reduced to their lowest terms using the common gcd, they are:
That is, these fractions when expressed in lowest terms, have the same denominator. This is true for cyclic permutations of any integer.
An integral multiplier refers to the multiplier n being an integer:
It is necessary for F to be coprime to 10 in order that 1⁄F is a repeating decimal without any preceding non-repeating digits (see multiple sections of Repeating decimal). If there are digits not in a period, then there is no corresponding solution.
For these two cases, multiples of X, i.e. (j X) are also solutions provided that the integer i satisfies the condition n j⁄F < 1. Most often it is convenient to choose the smallest F that fits the above. The solutions can be expressed by the formula:
To exclude integers that begin with zeros from the solutions, select an integer j such that j⁄F > 1⁄10, i.e. j > F⁄10.
There is no solution when n > F.
An integer X shift left cyclically by k positions when it is multiplied by a fraction n⁄s. X is then the repeating digits of s⁄F, whereby F is F_{0} = s 10^{k} - n, or a factor of F_{0}; and F must be coprime to 10.
For this third case, multiples of X, i.e. (j X) are again solutions but the condition to be satisfied for integer j is that n j⁄F < 1. Again it is convenient to choose the smallest F that fits the above.
The solutions can be expressed by the formula:
To exclude integers that begin with zeros from the solutions, select an integer j such that j s⁄F > 1⁄10, i.e. j > F⁄10s.
Again if j s⁄F > 1, there is no solution.
The direct algebra approach to the above cases integral multiplier lead to the following formula:
A long division of 1 by 7 gives:
0.142857... 7 ) 1.000000 .7 3 28 2 14 6 56 4 35 5 49 1
At the last step, 1 reappears as the remainder. The cyclic remainders are {1, 3, 2, 6, 4, 5}. We rewrite the quotients with the corresponding dividend/remainders above them at all the steps:
Dividend/Remainders 1 3 2 6 4 5 Quotients 1 4 2 8 5 7
and also note that:
By observing the remainders at each step, we can thus perform a desired cyclic permutation by multiplication. E.g.,
In this manner, cyclical left or right shift of any number of positions can be performed.
Less importantly, this technique can be applied to any integer to shift cyclically right or left by any given number of places for the following reason:
An integer X shift cyclically right by k positions when it is multiplied by an integer n. Prove its formula.
Proof
First recognize that X is the repeating digits of a repeating decimal, which always possesses cyclic behavior in multiplication. The integer X and its multiple n X then will have the following relationship:
This completes the proof.
An integer X shift cyclically left by k positions when it is multiplied by an integer n. Prove its formula.
Proof
First recognize that X is the repeating digits of a repeating decimal, which always possesses a cyclic behavior in multiplication. The integer X and its multiple n X then will have the following relationship:
which represents the results after left cyclical shift of k positions.
This completes the proof. The proof for non-integral multiplier such as n⁄s can be derived in a similar way and is not documented here.
The permutations can be:
Main article: parasitic number |
When a parasitic number is multiplied by n, not only it exhibits the cyclic behavior but the permutation is such that the last digit of the parasitic number now becomes the first digit of the multiple. For example, 102564 x 4 = 410256. Note that 102564 is the repeating digits of 4⁄39 and 410256 the repeating digits of 16⁄39.
An integer X shift right cyclically by double positions when it is multiplied by an integer n. X is then the repeating digits of 1⁄F, whereby F = n × 10^{2} - 1; or a factor of it; but excluding values for which 1⁄F has a period length dividing 2 (or, equivalently, less than 3); and F must be coprime to 10.
Most often it is convenient to choose the smallest F that fits the above.
The following multiplication moves the last two digits of each original integer to the first two digits and shift every other digits to the right:
Multiplier n | Solution | Represented by | Other Solutions |
---|---|---|---|
2 | 0050251256 2814070351 7587939698 4924623115 5778894472 3618090452 2613065326 6331658291 4572864321 608040201 | 1⁄199 x 2 = 2⁄199
period = 99 i.e. 99 repeating digits. |
2⁄199, 3⁄199, ..., 99⁄199 |
3 | 0033444816 0535117056 8561872909 6989966555 1839464882 9431438127 090301 | 1⁄299 x 3 = 3⁄299
period = 66 299 = 13×23 |
2⁄299, 3⁄299, ..., 99⁄299
some special cases are illustrated below |
3 | 076923 | 1⁄13 x 3 = 3⁄13
period = 6 |
2⁄13, 3⁄13, 4⁄13 |
3 | 0434782608 6956521739 13 | 1⁄23 x 3 = 3⁄23
period = 22 |
2⁄23, 3⁄23, ..., 7⁄23 |
4 | 0025062656 64160401 | 1⁄399 x 4 = 4⁄399
period = 18 399 = 3×7×19 |
2⁄399, 3⁄399, ..., 99⁄399
some special cases are illustrated below |
4 | 142857 | 1⁄7 x 4 = 4⁄7
period = 6 |
- |
4 | 0526315789 47368421 | 1⁄19 x 4 = 4⁄19
period = 18 |
2⁄19, 3⁄19, 4⁄19 |
5 | (a cyclic number with a period of 498) | 1⁄499 x 5 = 5⁄499
499 is a full reptend prime |
2⁄499, 3⁄499, ..., 99⁄499 |
Note that:
There are many other possibilities.
Problem: An integer X shift left cyclically by single position when it is multiplied by 3. Find X.
Solution: First recognize that X is the repeating digits of a repeating decimal, which always possesses some interesting cyclic behavior in multiplications. The integer X and its multiple then will have the following relationship:
This yields the results that:
The other solution is represented by 2⁄7 x 3 = 6⁄7:
There are no other solutions ^{[1]} because:
However, if the multiplier is not restricted to be an integer (though ugly), there are many other solutions from this method. E.g., if an integer X shift right cyclically by single position when it is multiplied by 3⁄2, then 3 shall be the next remainder after 2 in a long division of a fraction 2⁄F. This deduces that F = 2 x 10 - 3 = 17, giving X as the repeating digits of 2⁄17, i.e. 1176470588235294, and its multiple is 1764705882352941.
The following summarizes some of the results found in this manner:
Multiplier n⁄s | Solution | Represented by | Other Solutions |
---|---|---|---|
1⁄2 | 105263157894736842 | 2⁄19 × 1⁄2 = 1⁄19
A 2-parasitic number |
Other 2-parasitic numbers:
4⁄19, 6⁄19, 8⁄19, 10⁄19, 12⁄19, 14⁄19, 16⁄19, 18⁄19 |
3⁄2 | 1176470588235294 | 2⁄17 × 3⁄2 = 3⁄17 | 4⁄17, 6⁄17, 8⁄17, 10⁄17 |
7⁄2 | 153846 | 2⁄13 × 7⁄2 = 7⁄13 | - |
9⁄2 | 18 | 2⁄11 × 9⁄2 = 9⁄11 | - |
7⁄3 | 1304347826086956521739 | 3⁄23 × 7⁄3 = 7⁄23 | 6⁄23, 9⁄23, 12⁄23, 15⁄23, 18⁄23, 21⁄23 |
19⁄4 | 190476 | 4⁄21 × 19⁄4 = 19⁄21 | - |
An integer X shift left cyclically by double positions when it is multiplied by an integer n. X is then the repeating digits of 1⁄F, whereby F is R = 10^{2} - n, or a factor of R; excluding values of F for which 1⁄F has a period length dividing 2 (or, equivalently, less than 3); and F must be coprime to 10.
Most often it is convenient to choose the smallest F that fits the above.
The following summarizes some of the results obtained in this manner, where the white spaces between the digits divide the digits into 10-digit groups:
Multiplier n | Solution | Represented by | Other Solutions |
---|---|---|---|
2 | 142857 | 1⁄7 × 2 = 2⁄7 | 2⁄7, 3⁄7 |
3 | 0103092783 5051546391 7525773195 8762886597 9381443298 9690721649 4845360824 7422680412 3711340206 185567 | 1⁄97 x 3 = 3⁄97 | 2⁄97, 3⁄97, 4⁄97, 5⁄97, ...., 31⁄97, 32⁄97 |
4 | No solution | - | - |
5 | 0526315789 47368421 | 1⁄19 x 5 = 5⁄19 | 2⁄19, 3⁄19 |
6 | 0212765957 4468085106 3829787234 0425531914 893617 | 1⁄47 x 6 = 6⁄47 | 2⁄47, 3⁄47, 4⁄47, 5⁄47, 6⁄47, 7⁄47 |
7 | 0322580645 16129 | 1⁄31 x 7 = 7⁄31 | 2⁄31, 3⁄31, 4⁄31
1⁄93, 2⁄93, 4⁄93, 5⁄93, 7⁄93, 8⁄93, 10⁄93, 11⁄93, 13⁄93 |
8 | 0434782608 6956521739 13 | 1⁄23 x 8 = 8⁄23 | 2⁄23 |
9 | 076923 | 1⁄13 x 9 = 9⁄13 | 1⁄91, 2⁄91, 3⁄91, 4⁄91, 5⁄91, 6⁄91, 8⁄91, 9⁄91, 10⁄91 |
10 | No solution | - | - |
11 | 0112359550 5617977528 0898876404 4943820224 7191 | 1⁄89 x 11 = 11⁄89 | 2⁄89, 3⁄89, 4⁄89, 5⁄89, 6⁄89, 7⁄89, 8⁄89 |
12 | No solution | - | - |
13 | 0344827586 2068965517 24137931 | 1⁄29 x 13 = 13⁄29 | 2⁄29
1⁄87, 2⁄87, 4⁄87, 5⁄87, 6⁄87 |
14 | 0232558139 5348837209 3 | 1⁄43 x 14 = 14⁄43 | 2⁄43, 3⁄43 |
15 | 0588235294 117647 | 1⁄17 x 15 = 15⁄17 | - |
In duodecimal system, the transposable integers are: (using inverted two and three for ten and eleven, respectively)
Multiplier n | Smallest solution such that the multiplication moves the last digit to left | Digits | Represented by | Smallest solution such that the multiplication moves the first digit to right | Digits | Represented by |
---|---|---|---|---|---|---|
2 | 06316948421 | Ɛ | 1⁄1Ɛ x 2 = 2⁄1Ɛ | 2497 | 4 | 1⁄5 x 2 = 2⁄5 |
3 | 2497 | 4 | 1⁄5 x 3 = 3⁄5 | no solution | ||
4 | 0309236ᘔ8820 61647195441 | 1Ɛ | 1⁄3Ɛ x 4 = 4⁄3Ɛ | no solution | ||
5 | 025355ᘔ94330 73ᘔ458409919 Ɛ7151 | 25 | 1⁄4Ɛ x 5 = 5⁄4Ɛ | 186ᘔ35 | 6 | 1⁄7 x 5 = 5⁄7 |
6 | 020408142854 ᘔ997732650ᘔ1 83469163061 | 2Ɛ | 1⁄5Ɛ x 6 = 6⁄5Ɛ | no solution | ||
7 | 01899Ɛ864406 Ɛ33ᘔᘔ1542391 374594930525 5Ɛ171 | 35 | 1⁄6Ɛ x 7 = 7⁄6Ɛ | no solution | ||
8 | 076Ɛ45 | 6 | 1⁄17 x 8 = 8⁄17 | no solution | ||
9 | 014196486344 59Ɛ9384Ɛ26Ɛ5 33040547216ᘔ 1155Ɛ3Ɛ12978 ᘔ3991 | 45 | 1⁄8Ɛ x 9 = 9⁄8Ɛ | no solution | ||
ᘔ | 08579214Ɛ364 29ᘔ7 | 14 | 1⁄15 x ᘔ = ᘔ⁄15 | no solution | ||
Ɛ | 011235930336 ᘔ53909ᘔ873Ɛ3 25819Ɛ997505 5Ɛ54ᘔ3145ᘔ42 694157078404 491Ɛ1 | 55 | 1⁄ᘔƐ x Ɛ = Ɛ⁄ᘔƐ | no solution |
Note that the “Shifting left cyclically by single position” problem has no solution for the multiplier less than 12 except 2 and 5, the same problem in decimal system has no solution for the multiplier less than 10 except 3.